6. Simplify the ff. (a) $$ \frac{x}{4}+\frac{3}{7} $$ b) $$ \frac{2 x}{4+x}+\frac{x}{3} $$ (3) $$ \frac{a+1}{b} imes \frac{b^{2}}{a^{2}-1} $$

Question

6. Simplify the ff. (a) $$ \frac{x}{4}+\frac{3}{7} $$ b) $$ \frac{2 x}{4+x}+\frac{x}{3} $$ (3) $$ \frac{a+1}{b} 	imes \frac{b^{2}}{a^{2}-1} $$

UpStudy AI · Accepted Answer

(a) Combine the fractions by finding a common denominator. The least common denominator (LCD) of 4 and 7 is 28.

$$
\frac{x}{4}+\frac{3}{7} = \frac{7x}{28}+\frac{12}{28} = \frac{7x+12}{28}
$$

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(b) Write the sum with the common denominator $$3(4+x)$$.

$$
\frac{2 x}{4+x}+\frac{x}{3} = \frac{2x\cdot 3}{3(4+x)}+\frac{x(4+x)}{3(4+x)} = \frac{6x + x(4+x)}{3(4+x)}
$$

Expand the numerator:

$$
6x + x(4+x)= 6x + 4x + x^2 = x^2+10x
$$

Thus, the expression simplifies to:

$$
\frac{x^2+10x}{3(4+x)} = \frac{x(x+10)}{3(4+x)}
$$

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(3) Factor the difference of squares in the denominator.

$$
a^2-1=(a-1)(a+1)
$$

Then multiply:

$$
\frac{a+1}{b} \times \frac{b^{2}}{a^{2}-1} = \frac{a+1}{b}\times \frac{b^2}{(a-1)(a+1)}
$$

Cancel the common factor $$(a+1)$$ and one factor of $$b$$:

$$
= \frac{b}{a-1}
$$
(a) $$ \frac{7x + 12}{28} $$ (b) $$ \frac{x(x + 10)}{3(4 + x)} $$ (3) $$ \frac{b}{a - 1} $$

6. Simplify the ff. (a) \( \frac{x}{4}+\frac{3}{7} \) b) \( \frac{2 x}{4+x}+\frac{x}{3} \) (3) \( \frac{a+1}{b} \times \frac{b^{2}}{a^{2}-1} \)